Problem source

In this question, $\lfloor x \rfloor$ denotes the greatest integer that  is less than or equal to $x$, so that  $\lfloor 2.9 \rfloor = 2 = \lfloor 2.0 \rfloor$
and $\lfloor -1.5 \rfloor = -2$.
The function $\f$ is defined, for $x\ne0$, by $\f(x) = \dfrac{\lfloor x \rfloor}{x}\,$.
\begin{questionparts}
\item Sketch the graph of $y=\f(x)$ for $-3\le x \le 3$ (with $x\ne0$).
\item By considering the line $y= \frac7{12}$ on your graph, or otherwise, solve the equation $\f(x) = \frac7 {12}\,$.
Solve also the equations   $\f(x) =\frac{17}{24}$ and $\f(x) = \frac{4 }{3 }\,$.
\item
Find the largest root of the equation $\f(x) =\frac9{10}\,$.
\end{questionparts}
Give necessary and sufficient conditions, in the form of inequalities, for the equation $\f(x) =c$ to have exactly $n$ roots, where $n\ge1$.

Solution source

\begin{questionparts}
\item 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){floor(#1)/(#1)^2)};
    \def\xl{-4};
    \def\xu{4};
    \def\yl{-4};
    \def\yu{4};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{1}
    \pgfmathsetmacro{\yscale}{\xrange/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=-3:-2.01, samples=100] 
            plot (\x, {-3/\x});
        \draw[thick, blue, smooth, domain=-2:-1.01, samples=100] 
            plot (\x, {-2/\x});
        \draw[thick, blue, smooth, domain=-.99:-0.05, samples=100] 
            plot (\x, {-1/\x});  
        \draw[thick, blue, smooth, domain=2:3, samples=100] 
            plot (\x, {2/\x});
        \draw[thick, blue, smooth, domain=1:1.99, samples=100] 
            plot (\x, {1/\x});
        \draw[thick, blue, smooth, domain=0.05:1, samples=100] 
            plot (\x, {\functionf(\x)});  
    \end{scope}

    \draw[red, dashed] (-4, {7/12}) -- (4,{7/12});
    \draw[green, dashed] (-4, {17/24}) -- (4,{17/24});
    
    \end{tikzpicture}
\end{center}


\item Notice that there are no solutions when $x < 0$ since $f(x) \geq 1$ in that region.

Suppose $x = n + \epsilon, 0 < \epsilon < 1$, then $f(x) = \frac{n}{n+\epsilon}$, ie $12n = 7n + 7 \epsilon \Rightarrow 5 n = 7\epsilon \Rightarrow \epsilon = \frac{5}{7}n \Rightarrow n < \frac75$, so $n = 1 ,\epsilon = \frac57, x = \frac{12}5$.

\begin{align*}
&& \frac{17}{24} &= f(x) \\
\Rightarrow && 17n + 17 \epsilon &= 24 n \\
\Rightarrow && 17 \epsilon &= 7 n \\
\Rightarrow && n &< \frac{17}{7} \\
\Rightarrow && n &= 1, 2 \\
\Rightarrow && x &= \frac{24}{17}, \frac{48}{17}
\end{align*}.

For $f(x) = \frac{4}{3}$ we notice that $x < 0$, so let $x = -n +\epsilon$, ie 

\begin{align*}
&& \frac43 &= f(x) \\
\Rightarrow && \frac43 &= \frac{-n}{-n+\epsilon} \\
\Rightarrow && 4\epsilon &= n \\
\Rightarrow && n &= 1,2,3 \\
\Rightarrow && x &= -\frac{5}{4}, -\frac{3}{2}, -\frac{9}{4}
\end{align*}
\item \begin{align*}
&& \frac9{10} &= f(x) \\
\Rightarrow && 9n + 9 \epsilon &= 10 n \\
\Rightarrow && 9 \epsilon &= n \\
\Rightarrow && n < 9
\end{align}
So largest will be when $n = 8, \epsilon = \frac{8}{9}$, ie $\frac{80}{9}$
\end{questionparts}

If $c < 1$
\begin{align*}
&& c &= \frac{k}{k + \epsilon} \\
\Rightarrow && \frac{c}{1-c} \epsilon &= k
\end{align*}

For this to have exactly $n$ solutions, we need $n < \frac{c}{1-c} \leq n+1$.

If $c > 1$
\begin{align*}
&& c &= \frac{-k}{-k+\epsilon} \\
\Rightarrow && c \epsilon &= (c-1) k \\
\Rightarrow && \frac{c}{c-1} \epsilon &= k
\end{align*}
Therefore for there to be exactly $n$ solutions we need $n < \frac{c}{c-1} \leq n+1$